By considering the binomial expansions of \(\left(z+\frac{1}{z}\right)^{5}\) and \(\left(z-\frac{1}{z}\right)^{5}\), where \(z=\cos \theta+\mathrm{i} \sin \theta\), use de Moivre's theorem to show that
\[\tan ^{5} \theta=\frac{\sin 5 \theta-a \sin 3 \theta+b \sin \theta}{\cos 5 \theta+a \cos 3 \theta+b \cos \theta}\]
where \(a\) and \(b\) are integers to be determined.